Chapter 11, Problem 23P

(a)

To determine

To Find: The period and frequency.

Answer to Problem 23P

The period of the motion is $0.49\text{ s}$ and the frequency of the motion is $2.04\text{ Hz}$ .

Explanation of Solution

Given:

Mass, $m=0.755\text{ kg}$

Spring constant, $k=124\text{ N/m}$

Initial speed, $v_i=2.96\text{ m/s = }{v}_{\max }$

Formula used:

At t = 0, time period (T) is obtained by the following formula,

  $T=2\pi \times \sqrt{\frac{m}{k}}$.................(1)

Here $m$ is the mass of the spring,

  $k$ is the stiffness of the spring,

Write the expression for the frequency.

  $f\text{ = }\frac{\text{1}}{T}$.................(2)

Here, $T$ is the time period of the oscillation of spring.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}T=2\pi \times \sqrt{\frac{0.755}{124}}=0.49\text{ s}\\ f=\frac{1}{0.49}=2.04\text{ Hz}\end{array}$

Conclusion:

Thus, the period of the motion is $0.49\text{ s}$ and the frequency of the motion is $2.04\text{ Hz}$ .

(b)

To determine

To Find: The amplitude.

Answer to Problem 23P

The amplitude of the motion is $0.231\text{ m}$

Explanation of Solution

Given:

Mass, $m=0.755\text{ kg}$

Spring constant, $k=124\text{ N/m}$

Initial speed, $v_i=2.96\text{ m/s = Maximum velocity }{v}_{\max }$

Formula used:

Amplitude of the motion is determined by the following formula,

  $V_{\max }=\omega \times A$.................(3)

Here, $V_{\max }$ is the maximum velocity,

  $\omega =\sqrt{\frac{k}{m}}$.................(4)

Here $m$ is the mass of the spring,

  $k$ is the stiffness of the spring,

From equation 3 and equation 4.

  $V_{\max }=\sqrt{\frac{k}{m}}\times A$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{c}{V}_{\max }=\sqrt{\frac{k}{m}}\times A\\ A=V_{\max }\times \sqrt{\frac{m}{k}}\\ =2.96\times \sqrt{\frac{0.755}{124}}\\ =0.231\text{ m }\end{array}$

Conclusion:

Thus, the amplitude of the motion is $0.231\text{ m}$ .

(c)

To determine

To Find: The maximum acceleration.

Answer to Problem 23P

The maximum acceleration is $37.9{\text{ m/s}}^{\text{2}}$ .

Explanation of Solution

Given:

Mass $m=0.755\text{ kg}$

Amplitude $A=0.231\text{ m}$

Spring constant $k=124\text{ N/m}$

Formula used:

The maximum acceleration is determined by the following formula,

  $a_{\max }=A\times {\omega }^2$.................(5)

Here $A$ is the amplitude of the vibration in the spring,

  $\omega$ is the angular velocity.

  $\omega =\sqrt{\frac{k}{m}}$.................(6)

Here, $m$ is the mass of the spring,

  $k$ is the stiffness of the spring,

On comparing equation (5) and equation (6)

  $a_{\max }=A\times \frac{k}{m}$Calculation:

Substitute the given values in the above equation,

  $a_{\max }=A\times \frac{k}{m}=0.231\times \frac{124}{0.755}=37.9{\text{ m/s}}^{\text{2}}$

Conclusion:

Thus, the maximum acceleration is $37.9{\text{ m/s}}^{\text{2}}$

(d)

To determine

To Find: The position as a function of time.

Answer to Problem 23P

The position as a function of time is $x=0.231\sin (12.82)t$ .

Explanation of Solution

Given:

Amplitude $A=0.231\text{ m}$

Frequency $f=2.04\text{ Hz}$

Formula used:

Mass started at equilibrium position at x = 0, so the displacement (position) function with respect to time is given by sine function.

  $x=A\sin (\omega t)$

Here, $A$ is the amplitude of the vibration,

  $\omega$ is the angular velocity,

  $t$ is the time period.

  $\omega =2\pi f$

  $f$ is the frequency of the vibration.

Calculation:

Substitute the given values,

  $\begin{array}{l}x=A\sin (\omega t)\\ \omega =2\pi f\\ x=0.231\sin (2\pi \times 2.04\times t)\\ x=0.231\sin (12.82)t\end{array}$

Conclusion:

Thus, equation of displacement as a function of time is $x=0.231\sin (12.82)t$

(e)

To determine

To Find: The total energy.

Answer to Problem 23P

The total energy of the object is $3.31\text{ J}$

Explanation of Solution

Given:

Maximum velocity $V_{\max }=2.96\text{ m/s}$

Mass $m=0.755\text{ kg}$

Formula used:

The kinetic energy of an object is the total energy of an object at an equilibrium position, which is given by following formula,

  $E_{\max }\text{ = }\frac{1}{2}\times m\times v_{\max }{}^2$

Here, $v_{\max }$ is the maximum velocity,

  $m$ is the mass of the object.

Calculation:

Substitute the given values,

  $E_{\max }=\frac{1}{2}\times 0.755\times {(2.96)}^2=3.31\text{ J}$

Conclusion:

Thus, the total energy of the object is $3.31\text{ J}$ .